Question

Find the inverse of $A=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$ .

Answer 1

$A=\left[\begin{array}{ccc}cos\theta & -sin\theta & 0\\ sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right]$

Now , $\left|A\right|$ = $1(co{s}^2\theta +si{n}^2\theta )$ = $1$

adj A = $\left[\begin{array}{ccc}cos\theta & sin\theta & 0\\ -sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right]$

Hence,

Inverse of A = $A^{-1}$ = $\frac{1}{\left|A\right|}adjA$ = $\left[\begin{array}{ccc}cos\theta & sin\theta & 0\\ -sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right]$